Question

if two zeroes of the polynomial xbiquartic -6xcubic-26x square+138x-35 are 2±√3.find the other zeroes​

Answer 1

Answer:

Let p(x) =x4+6x3-26x2+138x-35 Given roots,2+-√3 . Since x=2+√3 is a zero. x-(2+√3) is a factor, x-2-√3 is a factor. Therefore x=2-√3 is a factor. x-(2-√3) is a factor, x-2+√3 is a factor. Since both are factors. Hence, (x-2-√3).(x-2+√3) =[(x-2)-√3] . [(x-2)+√3] = (x-2)2- (√3)2 =x2+4-4x-3 = x2-4x+1. Therefore, x2-4x+1 is a factor of p(x). x4-6x3-26x2+138x-35 ÷x2-4x+1 =x2-2x-25. solve the x2-2x-25 ,x2-7x+5x-35=0 , x(x-7)+5(x-7)=0 , (x+5)(x-7)=0. So, x=-5 and x=7 are roots. Hence the zeros of p(x) are 2+√3 , 2-√3 , -5 and 7.

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Step-by-step explanation: